1. Capillary rise formula

For a liquid rising in a capillary tube,

$h = \frac{2T\cos\theta}{\rho g r}$

where:

• $T = 0.05\,\text{N m}^{-1}$
• $r = 0.15\,\text{mm} = 1.5\times 10^{-4}\,\text{m}$
• $\rho = 667\,\text{kg m}^{-3}$
• $g = 10\,\text{m s}^{-2}$
• $\theta$ = angle of contact

2. Find the angle of contact

The two tangents drawn at the liquid-glass interfaces on opposite sides make an angle of $60^\circ$ with each other.

For a capillary meniscus, each tangent makes an angle $\theta$ with the wall inside the liquid. By geometry, the angle between the two tangents is

$180^\circ - 2\theta$

Given,

$180^\circ - 2\theta = 60^\circ$

$2\theta = 120^\circ$

$\theta = 60^\circ$

Hence,

$\cos\theta = \cos 60^\circ = \frac{1}{2}$

3. Substitute into the formula

$h = \frac{2(0.05)(1/2)}{667\times 10\times 1.5\times 10^{-4}}$

The numerator is:

$2\times 0.05\times \frac{1}{2} = 0.05$

The denominator is:

$667\times 10\times 1.5\times 10^{-4} = 6670\times 1.5\times 10^{-4} = 1.0005$

Thus,

$h = \frac{0.05}{1.0005} \approx 0.04997\,\text{m}$

So,

$h \approx 0.050\,\text{m}$

4. Match with options

This is closest to:

Option A: $0.049\,\text{m}$

5. Comparison with stored answer

The stored correct answer is B, but the calculation clearly gives A.

Therefore, the stored answer appears to be incorrect.